Chapter 10.2, Problem 5IP

To determine

To Find: To find the measures of variability and outliers if any of the given data.

Answer to Problem 5IP

The answer is

  $\text{Range}=1.05,\text{Inter-quartile}\text{range}=0.2025,\text{Variance}=0.0793\text{and}\text{Standard}\text{deviation}=0.282$

and the outlier is 1.25

Explanation of Solution

Given information:Sales of popcorn at movie time theatre in different years.

Formula used:

  $\begin{array}{l}\text{For}\text{Range}=\text{Highest}\text{value}-\text{Lowest}\text{value}\\ \text{For Inter-Quartile Range}={\text{Q}}_{\text{3}}-{\text{Q}}_{\text{1}}\\ \text{For Variance}=\frac{{\displaystyle \sum _{i=1}^n{(x_i)}^2}}{n}-{\overline{x}}^2\\ \text{For Standard Deviation}=\sqrt{\text{Variance}}\\ \text{For}\text{Outliers}:\text{Data}\text{Value}<\text{}{\text{Q}}_{\text{1}}-\text{1}\text{.5(IQR)}\text{or}\text{Data}\text{Value}>{\text{Q}}_{\text{3}}+\text{1}\text{.5(IQR)}\end{array}$

Calculation:

Since, the given data is 0.66, 0.43, 1.25, 0.2, 0.53, 0.6, 0.58, and 0.48.

Arranging the data in ascending order

0.2, 0.43, 0.48, 0.53, 0.58, 0.6, 0.66, 1.25

Now, the range of the given data is

  $\text{Range}=1.25-0.2=1.05$

Also, first and third quartile are

  $\begin{array}{l}{\text{Q}}_{\text{1}}={\left(\frac{n+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{1}}={\left(\frac{8+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{1}}={\left(\frac{9}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{1}}={(}^{2.25}\text{observation}\\ {\text{Q}}_1=2^{\text{nd}}\text{observation}+0.25(3^{\text{rd}}\text{observation}-2^{\text{nd}}\text{observation})\\ {\text{Q}}_{\text{1}}=0.43+0.25(0.48-0.43)\\ {\text{Q}}_{\text{1}}=0.43+0.0125=0.4425\end{array}$

and

  $\begin{array}{l}{\text{Q}}_{\text{3}}=3{\left(\frac{n+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{3}}=3{\left(\frac{8+1}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{3}}=3{\left(\frac{9}{4}\right)}^{\text{th}}\text{observation}\\ {\text{Q}}_{\text{3}}={(}^{6.75}\text{observation}\\ {\text{Q}}_3=6^{\text{th}}\text{observation}+0.75(7^{\text{th}}\text{observation}-6^{\text{th}}\text{observation})\\ {\text{Q}}_{\text{3}}=0.6+0.75(0.66-0.6)\\ {\text{Q}}_{\text{3}}=0.6+0.045=0.645\end{array}$

Thus, Inter-quartile range is $Q_3-Q_1=0.645-0.4425=0.2025$

Calculation of Mean and Variance

  $\begin{array}{l}\text{Mean}=\frac{\text{Sum}\text{of}\text{all}\text{observations}}{\text{Number}\text{of}\text{observations}}\\ \text{Mean}=\frac{0.2+0.43+0.48+0.53+0.58+0.6+0.66+1.25}{8}\\ \text{Mean}=\frac{4.73}{8}\\ \text{Mean}=0.59125\\ \text{Variance}=\frac{{\displaystyle \sum _{i=1}^n{(x_i)}^2}}{n}-{\overline{x}}^2,\text{where}\\ \frac{{\displaystyle \sum _{i=1}^n{x}_i^2}}{n}=\frac{{(0.2)}^2+{(0.43)}^2+{(0.48)}^2+{(0.53)}^2+{(0.58)}^2+{(0.6)}^2+{(0.66)}^2+{(1.25)}^2}{8}\\ \frac{{\displaystyle \sum _{i=1}^n{x}_i^2}}{n}=\frac{0.04+0.1849+0.2304+0.2809+0.3364+0.36+0.4356+1.5625}{8}\\ \frac{{\displaystyle \sum _{i=1}^n{x}_i^2}}{n}=\frac{3.4307}{8}=0.4288375\\ \text{Now,}\\ \text{Variance}=0.4288375-{(}^{0.59125}\\ \text{Variance}=0.4288375-0.3495766\\ \text{Variance}=0.0793\end{array}$

Calculation of standard deviation

  $\begin{array}{l}\text{Standard Deviation}=\sqrt{\text{Variance}}\\ \text{Standard}\text{Deviation}=\sqrt{0.0793}\\ \text{Standard}\text{Deviation}=0.282\end{array}$

Calculation of Outliers

  $\begin{array}{l}\text{Data}\text{Value}<{\text{Q}}_{\text{1}}-1.5\text{(Inter}\text{quartile}\text{range)}\\ \text{Data}\text{Value}<0.4425-1.5\times 0.2025\\ \text{Data}\text{Value}<0.4425-0.30375\\ \text{Data}\text{Value}<0.13875\end{array}$

Which is not exist in the given set of data, also

  $\begin{array}{l}\text{Data}\text{Value}>{\text{Q}}_{\text{3}}+1.5\text{(Inter}\text{quartile}\text{range)}\\ \text{Data Value}>0.645+1.5\times 0.2025\\ \text{Data}\text{Value}>0.645+0.30375\\ \text{Data}\text{Value}>0.94875\end{array}$

Since, $1.25>0.94875$

Therefore, the outlier is 1.25

Hence, the required measure of variability is

  $\text{Range}=1.05,\text{Inter-quartile}\text{range}=0.2025,\text{Variance}=0.0793\text{and}\text{Standard}\text{deviation}=0.282$and the outlier is 1.25